IB Physics: Formula Derivations for the IB Diploma | IB 物理：Physics for the IB Diploma 公式推导

📚 IB Physics: Formula Derivations for the IB Diploma | IB 物理：Physics for the IB Diploma 公式推导

Formulas are the language of physics, but memorising them without understanding their origins can leave gaps in reasoning. In the IB Diploma programme, derivations reveal the logical links between fundamental principles and the equations we use to solve real problems. This article walks through the derivations of essential formulas in mechanics, thermal physics, waves, electricity, and quantum phenomena, helping you build conceptual depth and exam confidence.

公式是物理学的语言，但只记忆而不理解其来源会留下推理漏洞。在 IB 文凭课程中，公式推导揭示了基本原理与我们解题所用方程之间的逻辑联系。本文逐步推演力学、热学、波动、电学和量子现象中的核心公式，帮助你建立概念深度与应试信心。

1. Kinematic Equations | 运动学方程推导

For uniformly accelerated motion, the average acceleration a is defined as the rate of change of velocity. If initial velocity is u, final velocity is v, and time taken is t, then:

对于匀加速运动，平均加速度 a 被定义为速度的变化率。若初速度为 u，末速度为 v，所用时间为 t，则：

a = (v − u) / t

Rearranging immediately gives the first kinematic equation:

重新整理后立刻得到第一个运动学方程：

v = u + at    (1)

When acceleration is constant, the average velocity is the arithmetic mean of initial and final velocities:

当加速度恒定时，平均速度为初速度与末速度的算术平均值：

v_avg = (u + v) / 2

Displacement s equals average velocity multiplied by time:

位移 s 等于平均速度乘以时间：

s = ((u + v) / 2) × t

Substituting v from equation (1) yields the second equation:

代入方程 (1) 中的 v 得到第二个方程：

s = ut + ½at²    (2)

To obtain the equation that links velocity and displacement without time, solve equation (1) for t and substitute into the displacement expression:

为得到不显含时间的速度–位移关系，从方程 (1) 解出 t 并代入位移表达式：

t = (v − u) / a ⇒ s = ((u + v)/2) × ((v − u)/a)

Simplifying gives the third kinematic equation:

化简后得到第三个运动学方程：

v² = u² + 2as    (3)

These three equations are the foundation of projectile and linear motion analysis in IB Physics.

这三个方程是 IB 物理中抛体运动和直线运动分析的基础。

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2. Newton’s Second Law and Momentum | 牛顿第二定律与动量

Newton's original formulation of his second law relates force to the rate of change of momentum p.

牛顿最初表述的第二定律将力与动量 p 的变化率联系起来。

F = dp / dt

If the mass of the object is constant, p = mv, so this simplifies to the familiar form:

若物体的质量恒定，p = mv，则上式简化为常见的形式：

F = m dv/dt = ma

The impulse J delivered by a net force over a time interval is the integral of the force, which equals the change in momentum:

净力在一段时间间隔内提供的冲量 J 是力对时间的积分，等于动量的变化量：

J = ∫ F dt = Δp = mv − mu

0, so total momentum remains constant. This underpins collision and explosion problems across the IB syllabus.

由此可导出动量守恒定律：若系统所受合外力为零，dp/dt = 0，因此总动量保持恒定。这是 IB 课程中碰撞与爆炸问题的基础。

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3. Work–Energy Theorem | 功能定理

When a constant net force F acts on an object over a displacement s in the direction of the force, the work done is W = F s.

当恒定的净力 F 沿力的方向对物体作用一段位移 s 时，所做的功为 W = F s。

Using Newton’s second law, F = m a, and the kinematic relation v² = u² + 2as, we can express a s as (v² − u²) / 2.

利用牛顿第二定律 F = m a 以及运动学关系 v² = u² + 2as，可将 a s 表示为 (v² − u²) / 2。

W = m a s = m × (v² − u²)/2 = ½mv² − ½mu²

The quantity ½mv² is defined as kinetic energy KE. Thus the net work done equals the change in kinetic energy:

½mv² 被定义为动能 KE。因此，净功等于动能的变化量：

W_net = ΔKE

This work–energy theorem allows us to solve many mechanics problems without needing to consider instantaneous accelerations.

功能定理使我们无需考虑瞬时加速度就能解决许多力学问题。

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4. Gravitational Potential Energy | 引力势能

To lift a body in a uniform gravitational field near the Earth’s surface, the work done against gravity is mgΔh, giving the familiar ΔU = mgΔh. For large distances, however, the gravitational force follows Newton’s law of universal gravitation, and we must integrate.

在地表附近的均匀引力场中提升物体，克服重力的功为 mgΔh，从而得到熟悉的 ΔU = mgΔh。然而，对于大距离，引力遵从万有引力定律，必须进行积分。

The force on a mass m due to a mass M is F = GMm / r², directed towards M. The work done by an external agent to move m from a distance r₁ to r₂ against this force is:

质量 m 受到质量 M 的引力为 F = GMm / r²，方向指向 M。外力将 m 从距离 r₁ 移动到 r₂ 克服引力所做的功为：

W = ∫_r₁^r₂ (GMm / r²) dr = GMm [ −1/r ]_r₁^r₂ = GMm (1/r₁ − 1/r₂)

Choosing the zero of potential energy at infinity (U → 0 as r → ∞), the potential energy at distance r is the work done to bring m from infinity to that point:

选取无穷远处势能为零（U → 0 当 r → ∞），则在距离 r 处的引力势能等于将 m 从无穷远移到该点所做的功：

U = − GMm / r

The negative sign confirms that gravity is an attractive force and work is done by the field as objects fall together. Near Earth, this expression reduces to U = mgh with g = GM / R².

负号表明引力是吸引力，物体相互靠近时引力场做正功。靠近地球表面，此表达式可近似为 U = mgh，其中 g = GM / R²。

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5. Simple Harmonic Motion | 简谐运动

Simple harmonic motion (SHM) occurs when the restoring force is directly proportional to the displacement from equilibrium and acts in the opposite direction: F = −k x.

当回复力与偏离平衡位置的位移成正比且方向相反，即 F = −k x 时，物体做简谐运动。

Applying Newton’s second law:

应用牛顿第二定律：

m a = −k x ⇒ a = − (k/m) x

Since acceleration is the second derivative of displacement, we look for a function whose second derivative is proportional to the negative of the function itself. This is satisfied by sine and cosine functions. If we set:

由于加速度是位移的二阶导数，我们寻找一个函数，其二阶导数正比于函数本身的负值。正弦和余弦函数满足此条件。若设：

x = A cos(ωt)

then the velocity and acceleration are:

则速度和加速度为：

v = dx/dt = −Aω sin(ωt)

a = d²x/dt² = −Aω² cos(ωt) = −ω² x

Comparing with a = − (k/m) x gives the angular frequency:

与 a = − (k/m) x 比较得到角频率：

ω = √k/m

The period T follows from ω = 2π / T:

周期 T 由 ω = 2π / T 得出：

T = 2π √m/k

This derivation applies to mass–spring systems and is generalised for any SHM scenario, including pendulums (where T = 2π √L/g for small angles).

此推导适用于弹簧振子系统，并可推广至任何简谐运动情形，包括小角度下的单摆 (T = 2π √L/g)。

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6. Kinetic Theory and the Ideal Gas Law | 分子动理论与理想气体定律

The microscopic model of an ideal gas considers N identical molecules moving randomly inside a cubic container of side L. Focus on one molecule with velocity component v_x perpendicular to a wall.

理想气体的微观模型考虑 N 个相同分子在边长为 L 的立方容器内随机运动。重点关注一个分子垂直于壁面的速度分量 v_x。

During a head-on elastic collision, the change in momentum of the molecule is 2m v_x. The time between successive collisions with the same wall is Δt = 2L / v_x. The average force exerted on the wall by this molecule is:

在迎面弹性碰撞中，该分子的动量变化为 2m v_x。与同一壁面连续两次碰撞的时间间隔为 Δt = 2L / v_x。此分子对壁面施加的平均力为：

F₁ = Δp / Δt = (2m v_x) / (2L / v_x) = m v_x² / L

Summing over all N molecules and using

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